Solve: 3sin⁡2x+4cos⁡2x−2cos⁡x+6sin⁡x−6=0 My Try 6sin⁡xcos⁡x+4(cos2x−sin2x)−2cos⁡x+6sin⁡x−6=0

Jason Yuhas

Jason Yuhas

Answered

2021-12-31

Solve: 3sin2x+4cos2x2cosx+6sinx6=0
My Try
6sinxcosx+4(cos2xsin2x)2cosx+6sinx6=0

Answer & Explanation

Timothy Wolff

Timothy Wolff

Expert

2022-01-01Added 26 answers

6sinxcosx+4(cos2xsin2x)2cosx+6sinx6=0
6sinxcosx9sin2xcos2x2cosx+6sinx1=0
(3sinxcosx)2+2(3sinxcosx)1=0
(3sinxcosx)22(3sinxcosx)+1=0
Philip Williams

Philip Williams

Expert

2022-01-02Added 39 answers

x=2y we get after some manipulation
4cos4y36cos2ysin2y+24cos3ysiny=0
4cos2y(cosy3siny)2=0
Vasquez

Vasquez

Expert

2022-01-09Added 457 answers

Abbreviating sinx and cosx to s and c, and writing sin2x=2sc and cos2x=2c21, the equation becomes
6sc+4(2c21)2c+6s6=0
Factoring out a 2 and grouping terms, we rewrite this as
3s(c+1)+4c2c5=0
But 4c2c5=(4c5)(c+1), so we have two solutions in s and c:
c=-1 and 3s+4c=5
Reverting to sinx and cosx and recognizing the Pythagorean triple 32+42=52, these become
cosx=1 and sin(x+θ)=1
where sinθ=45 and cosθ=35, e.g. θ=arcsin(45). So the solution set is
{π+2nπ|nZ}{π2arcsin(45)+2nπ|nZ}

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?